Answer:
76
Step-by-step explanation:
Answer:

Step-by-step explanation:
No it isn't, because:
![{1.08}^{ \frac{1}{5} } = { \frac{108}{100} }^{ \frac{1}{5} } = { \frac{27}{25} }^{ \frac{1}{5} } = \sqrt[5]{ \frac{27}{25} } = 1.01551](https://tex.z-dn.net/?f=%20%7B1.08%7D%5E%7B%20%5Cfrac%7B1%7D%7B5%7D%20%7D%20%20%3D%20%20%7B%20%5Cfrac%7B108%7D%7B100%7D%20%7D%5E%7B%20%5Cfrac%7B1%7D%7B5%7D%20%7D%20%20%3D%20%20%7B%20%5Cfrac%7B27%7D%7B25%7D%20%7D%5E%7B%20%5Cfrac%7B1%7D%7B5%7D%20%7D%20%20%3D%20%20%5Csqrt%5B5%5D%7B%20%5Cfrac%7B27%7D%7B25%7D%20%7D%20%20%3D%201.01551)
Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
![\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%261%262%5C%5C1%26-2%263%5Cend%7Barray%7D%5Cright%5D)
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
or
[
,
,
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[
,
,
]
In conclusion, the two unit vectors are;
[
,
,
]
and
[
,
,
]
<em>Hope this helps!</em>